Question

in a NaCl type structure of A and B , 4 atoms of A are removed from the edges and 4 atoms of B are removed from corners. the formula of the new compound will be

Answer 1

Answer:

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Since the structure is of NaCl , there are 4 A atoms and 4 B atoms in the unit cell. B atoms are present at thecorners and face centers and A atomsare present in octahedral voids. 4 corner B atoms had contribution4×81=21 to the unit cell. Thus 0.5 B atoms are removed and 4−0.

Answer 2

In a NaCl type structure of A and B, 4 atoms of A are removed from the edges and 4 atoms of B are removed from corners. The formula of the new compound will be A6B7.

• In a NaCl type structure of A and B, implies that the given AB is an FCC structure.
• Before removing face centred atoms, we have,
• The number of 'B' atoms = (8 × 1/8) + (6 × 1/2) = 1 + 3 = 4  
• The number of 'A' atoms = (12 × 1/4) + (1 × 1) = 3 + 1 = 4
• After removing of A and B atoms :
• The number of 'B' atoms = (4 × 1/8) + (6 × 1/2) = 1/2 + 3 = 7/2
• The number of 'A' atoms = (8 × 1/4) + (1 × 1) = 2 + 1 = 3
• So the formula of compound = A3B7/2
• Multiplying by 2, we get,
• Therefore, the formula of the new compound is = A6B7